In QED is angular momentum conserved at every order in Dyson series

[kof9595995]

I did some perturbative calculation earlier, and it seemed I have to impose angular momentum conservation by hand, or else I couldn't get the desired result. We know angular momentum must be conserved finally, but is it conserved at every order in Dyson series?

 

[Parlyne]

It ought to be. Angular momentum conservations is a consequence of rotational invariance, which, itself, is a subset of Lorentz invariance. The Dyson series is term-by-term Lorentz invariant; so, any perturbative calculation based on it ought to conserve angular momentum explicitly.

 

[tom.stoer]

The proof is technically rather involved.

One has to find a representation of U (or V) in the interaction picture and one has to construct the Lorentz generators and especially L (angular momentum) as well. Depending on the choice of the gauge this will be rather lengthy.

Then it should be clear that [L,V] = 0 is formally OK, but in principle a regularization for the operator products is required.

 

[kof9595995]

Thanks for the replies, and let me take some time thinking about it. And tom, where can I find the proof?

 

[tom.stoer]

And tom, where can I find the proof?

I don't know. I've calculated similar operator algebras years ago but not for the Dyson series.

 

[Bill_K]

Sorry, you are obviously not talking about the usual scattering of plane waves, since the orbital angular momentum of a plane wave is not a single eigenvalue. Are you thinking of expanding everything in terms of vector and spinor spherical harmonics??

 

[tom.stoer]

I am talking about

U(t,t_0) = \mathcal{T}\;\text{exp}\left[-i\int_{t_0}^t d\tau\,H_\text{int}(\tau)\right]

where U is the time volution operator, H_int is the interaction operator in the interaction picture and T is the time ordering

 

[Bill_K]

Tom, please be patient with me. I've been trying to understand what the complicating issue is here, why angular momentum conservation at the diagram level is not just an automatic consequence, and what would make it require detail proof. Take any Feynman diagram and rotate it. Each internal propagator is an invariant function. Each vertex is H_int, a scalar. Each external momentum goes p -> p' = Rp. Therefore the rotated diagram yields the same amplitude as the original. Doesn't this prove it?

 

[tom.stoer]

H_int is not a scalar but a piece of the 0th component of a four-vector. I think what one has to do is to show that each term in the Dyson series reproduces the correct Lorentz algebra; this is not trivial b/c the series contains the time-ordered exponential, whereas the usual Lorentz operator algebra is always calculated at constant = fixed time.

 

[kof9595995]

Sorry, you are obviously not talking about the usual scattering of plane waves, since the orbital angular momentum of a plane wave is not a single eigenvalue. Are you thinking of expanding everything in terms of vector and spinor spherical harmonics??

Actually I was motivated by the problem of 1S-state positronium decay(peskin , problem 5.4), I want to show triplet state cannot decay into 2 photons. This certainly can be done non-perturbatively by considering charge conjugation symmetry, but I want to show this by perturbation. I am able to show it in extreme nonrelativistic limit, where electron 4-momentum is given by (m,0,0,0), in this limit the M matrix simply vanish. Later I tried to show it in not-so-extreme non-relativsitic limit(in the sense that electron spin is still well defined), where p=(m,p1,p2,p3). Now the best hope is that M matrix is an odd function of 3-momentum p, i.e. M(p)+M(-p)=0, so that after weighted by 1S wavefunction(in momentum space) the amplitude will be 0. In proving M(p)+M(-p)=0, there's just some term I can't kill unless I restrict the polariztion vectors of final state photons to some special directions by angular momentum conservation, but this is not what I expected unless angular momentum conservation is not ingrained in the perturbation series, hence the question.
I can show my calculation here if anyone is patient enough to look at it, it's quite tedious.